SHM Question (1 Viewer)

pikachu975 · Jun 15, 2017

http://prntscr.com/fk07iy

There's no answers but I got C by differentiating 3 times then setting derivative of acceleration equal to 0 and got t = 3pi/2.

Not sure if there's a faster method, thanks!

InteGrand · Jun 15, 2017

pikachu975 said:
http://prntscr.com/fk07iy

There's no answers but I got C by differentiating 3 times then setting derivative of acceleration equal to 0 and got t = 3pi/2.

Not sure if there's a faster method, thanks!

$\noindent Let $\omega = \frac{1}{3}$, then by SHM facts, we know $\ddot{x} = -\omega^{2} x$. Thus acceleration $\ddot{x}$ is greatest when $x$ is least. On the given interval, this is actually when (and only when) $t = 0$, since $x(t) \geq 0$ for all $t \in [0,2\pi]$ with equality iff $t = 0$.$

$\noindent If the question meant when is the acceleration greatest in absolute value (which is probably the case based on the given options), then from the SHM differential equation, we can see that this is when $x$ is greatest in absolute value, and for $t$ in the given interval, this is when (and only when) $\frac{t}{3} = \frac{\pi}{2}$, i.e. $t = \frac{3\pi}{2}$.$

frog1944 · Jun 15, 2017

Why is acceleration the greatest when x is the least? If x was the least at say x=0, then acceleration would be zero?

Thanks

jathu123 · Jun 15, 2017

frog1944 said:
Why is acceleration the greatest when x is the least? If x was the least at say x=0, then acceleration would be zero?

Thanks

The acceleration is always 0 when x=centre of motion (cause a=-n^2(x-b)) . If x was the least at x=0, then the centre of motion would be somewhere above 0 (and acceleration would be 0 at that point)

SHM Question (1 Viewer)

pikachu975

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